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musickatia [10]

2 years ago

13

According to a report in USAToday, more and more parents are helping their young adult children get homes. Suppose eight persons

in a random sample of 40 young adults who recently purchased a home in Kentucky received help from their parents. You have been asked to construct a 95% confidence interval for the population proportion of all young adults in Kentucky who received help from their parents. What is the margin of error for a 95% confidence interval for the population proportion?

1 answer:

Inga [223]2 years ago

4 0

Answer:

a) 95% confidence interval for the population proportion of all young adults in Kentucky who received help from their parents.

( 0.0761 , 0.3239)

b) Margin of error = 0.1264.

Step-by-step explanation:

<u>Explanation</u>:-

Given '8' persons in a random sample of 40 young adults who recently purchased a home in Kentucky received help from their parents.

sample proportion of success $'p' = \frac{8}{40} = 0.2$

q = 1=p

q = 1-0.2 = 0.8

a)

95% confidence interval for the population proportion of all young adults in Kentucky who received help from their parents.

$(p-1.96\sqrt{\frac{pq}{n} } , p+1.96\sqrt{\frac{pq}{n} } )$

$(0.2-1.96\sqrt{\frac{0.2X0.8}{40} } , 0.2+1.96\sqrt{\frac{0.2X0.8}{40} } )$

(0.2 - 0.1239,0.2+0.1239)

( 0.0761 , 0.3239)

b) the margin of error for a 95% confidence interval for the population proportion.

For the 95% confidence interval ∝= 0.05 and zₐ = 1.96≅2.

$Margin of error = \frac{2\sqrt{pq} }{\sqrt{n} }$

$Margin of error = \frac{2\sqrt{0.2X0.8} }{\sqrt{40} }$

Margin of error for a 95% confidence interval for the population proportion.

Margin of error = 0.1264.

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Answer:

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Answer:

a) $\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) $\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) the steady state mass of the drug is 2000 mg

d) t ≅ 153.51 minutes

Step-by-step explanation:

From the given information;

At time t= 0

an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of 500

The inflow rate is 0.06 L/min.

Assume the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant.

The objective of the question is to calculate the following :

a) Write an initial value problem that models the mass of the drug in the blood for t ≥ 0.

From above information given :

$Rate _{(in)}= 500 \ mg/L \times 0.06 \ L/min = 30 mg/min$

$Rate _{(out)}=\dfrac{x}{4} \ mg/L \times 0.06 \ L/min = 0.015x \ mg/min$

Therefore;

$\dfrac{dx}{dt} = Rate_{(in)} - Rate_{(out)}$

with respect to x(0) = 0

$\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.

$\dfrac{dx}{dt} = -0.015(x - 2000)$

$\dfrac{dx}{(x - 2000)} = -0.015 \times dt$

By Using Integration Method:

$ln(x - 2000) = -0.015t + C$

$x -2000 = Ce^{(-0.015t)$

$x = 2000 + Ce^{(-0.015t)}$

However; if x(0) = 0 ;

Then

C = -2000

Therefore

$\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) What is the steady-state mass of the drug in the blood?

the steady-state mass of the drug in the blood when t = infinity

$\mathbf{x = 2000 - 2000e^{-0.015 \times \infty }}$

x = 2000 - 0

x = 2000

Thus; the steady state mass of the drug is 2000 mg

d) After how many minutes does the drug mass reach 90% of its stead-state level?

After 90% of its steady state level; the mas of the drug is 90% × 2000

= 0.9 × 2000

= 1800

Hence;

$\mathbf{1800 = 2000 - 2000e^{(-0.015t)}}$

$0.1 = e^{(-0.015t)$

$ln(0.1) = -0.015t$

$t = -\dfrac{In(0.1)}{0.015}$

t = 153.5056729

t ≅ 153.51 minutes

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